A normed field is a field $F$ together with a multiplicative norm $|\cdot|:F\to \R$. Given such a field, we form the ring $\Ocl_F := \{ x\in F : |x| \leq 1\}$, called the ring of integers of $F$. The ring of integers of $F$ is a local ring, with maximal ideal $\{ x\in F : |x| \le 1\}$.

A **perfectoid field** is a complete nonarchimedean normed field $(F,|\cdot|)$ that has residue characteristic $p > 0$, is not discretely valued, and such that ${\rm Fr}:\Ocl_F/p\to\Ocl_F/p$ is surjective, where ${\rm Fr}(x) = x^p$ is the Frobenius. An example of a perfectoid field is $\F_p((t))(t^{1/p^\infty})^\wedge$ — the field obtained from the Laurent series $\F_p((t))$ by appending all $p$-power roots of $t$ and then taking the completion.

Given a normed field $F$, define $\Ocl_{F^\flat}$ to be the inverse limit of the system $\cdots\to \Ocl_F/p\to\Ocl_F/p$ where the maps are the Frobenius, and $F^\flat$ to be the fraction field of $\Ocl_{F^\flat}$.

For $F = \Q_p$, then $F^\flat = \F_p$, and these two fields are very different from each other. If $F$ is perfectoid, then $F^\flat$ is also perfectoid and it turns out their Galois groups are isomorphic.

One can look at perfectoid fields through the notion of being deeply ramified: a normed field $F$ is called **deeply ramified** if for every finite extension $L/F$, the $\Ocl_L$-module of Kahler differentials $\Omega_{\Ocl_L/\Ocl_F}$ is zero. In this post, we will show that a perfectoid field is deeply ramified, following the proof written in Kedlaya's paper "New Methods for $(\Phi,\Gamma)$-Modules", with expanded details.

**Theorem.**If $L/F$ is a finite separable extension of normed fields then there exists an element $y\in \Ocl_L$ such that $y\Omega_{\Ocl_L/\Ocl_F} = 0$.

*Proof.*Consider the sequence $\Omega_F\to F\to L$. The first fundamental sequence for Kahler differentials is the exact sequence

$$\Omega_{F/\Ocl_F}\otimes_F L\to \Omega_{L/\Ocl_F}\to \Omega_{L/F}\to 0.$$

The first term is zero, being a localization of $\Omega_{\Ocl_F/\Ocl_F}$ and the third is zero because the extension $L/F$ is separable. Therefore

$$ 0 = \Omega_{L/\Ocl_F} = S^{-1}\Omega_{\Ocl_L/\Ocl_F}$$

where $S$ is the multiplicatively closed set $\Ocl_L – \{ 0\}$. Therefore, for each $dx\in\Omega_{\Ocl_L/\Ocl_F}$ there exists a $u\in\Ocl_L -\{ 0\}$ such that $udx = 0$. If we choose an basis $x_1,\dots,x_n\in\Ocl_L$ of $L$ as an $F$-vector space, then we can choose a $u\in \Ocl_L-\{0\}$ such that $ux_i = 0$ for all $i=1,\dots,n$. Furthermore, by choosing a $t\in\Ocl_L-\{0\}$ such that $t\Ocl_L\subseteq \sum_i \Ocl_F x_i$, we see that $y = tu$ is such that $y\Omega_{\Ocl_L/\Ocl_F} = 0$. QED

**Theorem.**A perfectoid field is deeply ramified.

*Proof.*Let $F$ be a perfectoid field and $L$ be a finite extension of $F$. Since perfectoid fields are either of characteristic zero or perfect, the extension $L/F$ is separable. Therefore, there exists a nonzero $y\in\Ocl_L$ such that $y\Omega_{\Ocl_L/\Ocl_F} = 0$.

Since $F$ is perfectoid and any finite extension of a perfectoid field is also perfectoid, the field $L$ is also perfectoid. In particular, the Frobenius $\Ocl_L/p\to\Ocl_L/p$ is surjective. Hence for $x\in \Ocl_L$ there exists $y,z\in\Ocl_L$ such that $x= y^p + pz$. In $\Omega_{\Ocl_L/\Ocl_F}$, we have

$$ dx = py^{p-1}dp + pdz$$

and hence $p^n\Omega_{\Ocl_L/\Ocl_F} = \Omega_{\Ocl_L/\Ocl_F}$ for all positive integers. Since $|p| \le 1$, there exists an $n$ such that $|p^n| \le y$, and so $y | p^n$. Since $y$ annihilates $\Omega_{\Ocl_L/\Ocl_F}$, we have

$$0 = p^n\Omega_{\Ocl_L/\Ocl_F} = \Omega_{\Ocl_L/\Ocl_F}$$. QED

Conversely, a complete nonarchimedean local field whose residue field has positive characteristic and is deeply ramified is perfectoid. A proof of this is part of Propositions 6.6.2 and 6.6.6 in Gabber and Ramero's strangely titled book "Almost Ring Theory", though unfortunately those proofs are not self-contained.

Hi Jason,

thank you for this interesting post!

I have a question about the proof of the first theorem, though: I don't understand why

0= S^{-1} \Omega_{o_L/o_F} implies that für every dx there is this element u such that udx=0.

Best wishes,

Outis

This is because by definition, if a localization $S^{-1}M = 0$ for some multiplicatively closed set $S$, then each element of the module $M$ is annihilated by some $s\in S$.